Question

exam 1 fall25
due sunday by 11:59pm points 100 submitting an external tool available until sep 21 at 11:59
exam 1 fall25
score: 45/100 answered: 13/18
question 14
solve: $\frac{4}{x + 11}=\frac{-88}{x^{2}-121}$
$x = $
enter your answer as a reduced fraction or an integer. enter dne for does not exist, ∞ for infinity
question help: message instructor
check answer

Explanation:

Step1: Factor the denominator

Note that $x^{2}-121=(x + 11)(x - 11)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. The equation $\frac{4}{x + 11}=\frac{-88}{x^{2}-121}$ becomes $\frac{4}{x + 11}=\frac{-88}{(x + 11)(x - 11)}$.

Step2: Multiply both sides by $(x + 11)(x - 11)$

Since $x
eq - 11$ (to avoid division by zero), we have $4(x - 11)=-88$.

Step3: Expand the left - hand side

$4x-44=-88$.

Step4: Add 44 to both sides

$4x=-88 + 44$, so $4x=-44$.

Step5: Solve for x

Divide both sides by 4: $x=\frac{-44}{4}=-11$. But if $x=-11$, the original equation has a denominator of zero in the left - hand side and the right - hand side. So the solution does not exist.

Answer:

DNE